Question: Let $k$ be a real number such that $k > 1$ and
\[\sum_{n=1}^{\infty} \frac{5n-1}{k^n} = \frac{13}{4}.\]Find $k.$
Explanation: Let $$S =\sum_{n=1}^{\infty} \frac{5n-1}{k^n} = \frac{4}{k} +   \frac{9}{k^2} +  \frac{14}{k^3} + \dotsb.$$Multiplying by $k$ gives us
$$kS = 4 + \frac{9}{k} +   \frac{14}{k^2} +  \frac{19}{k^3} + \dotsb.$$Subtracting the first equation from the second gives us
$$\begin{aligned}(k-1)S &= 4 + \frac{5}{k} +   \frac{5}{k^2} +  \frac{5}{k^3} + \dotsb \\
&= 4 + \frac{\frac{5}{k}}{1-\frac{1}{k}} \\
&= 4 + \frac{5}{k-1} \\
&= \frac{4k +1}{k-1}.
\end{aligned}$$Therefore,
$$S = \frac{4k +1}{(k-1)^2} = \frac{13}{4}.$$Rearranging gives,
$$16k + 4 = 13(k^2-2k+1).$$Bringing all the terms on one side gives us
$$13k^2-42k+9 = 0$$Factoring gives
$$(k-3)(13k-3) = 0.$$Hence, $k=3$ or $k= \frac{3}{13}$. Since we are told that $k > 1$ (and more importantly, the series converges), we have that $k = \boxed{3}.$